Arc Length

A piece of a curve lying between two points is called an arc. Its length comes from one idea: over an infinitesimal step the curve is virtually straight, so the Pythagorean theorem gives $d s = \sqrt{(d x)^{2} + (d y)^{2}}$ , and integrating $d s$ along the curve adds up all these infinitesimal pieces.

Item	Statement
The differential of arc length	$d s = \sqrt{(d x)^{2} + (d y)^{2}}$
Length of $y = f (x)$ , $a \leq x \leq b$	$L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$
Length of $x = h (y)$ , $c \leq y \leq d$	$L = \int_{c}^{d} \sqrt{{(\frac{d x}{d y})}^{2} + 1} d y$
When to switch to $y$	$d y / d x$ is undefined (a vertical tangent or a cusp)
Arc length function	\displaystyle s(x)=\int_{a}^{x}\sqrt{1+[f'(t)]^{2}}\,dt
Its derivative	s'(x)=\sqrt{1+[f'(x)]^{2}}>0, so $s$ is increasing
Rectifiable curve	one of finite length; guaranteed when \vert f'\vert on $[a, b]$
The whole trick	pick the algebra so that 1+(y')^{2} is a perfect square
Circle $x^{2} + y^{2} = R^{2}$	circumference $2 π R$
Catenary $y = a \cosh \frac{x}{a}$	length on $[0, b]$ is $a \sinh \frac{b}{a}$

The Differential of Arc Length

In this section, we want to find the length of the curve $y = f (x)$ from $x = a$ to $x = b$ . Like the previous sections of this chapter, we construct the formula by considering infinitesimals.

Let $s$ be the length of the arc from the fixed point $(a, f (a))$ to a variable point $(x, f (x))$ , as in the figure below. Assume $s$ increases by an infinitesimally small amount $d s$ , while $x$ and $y$ increase by $d x$ and $d y$ respectively. Because $d s$ is so small, this part of the curve is virtually straight, and the Pythagorean theorem gives

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{ds=\sqrt{(dx)^{2}+(dy)^{2}}}

The graph of y = f(x) with the arc s measured from the fixed point (a, f(a)) to a variable point (x, f(x)), and a small right triangle at that point whose horizontal leg is dx, whose vertical leg is dy, and whose hypotenuse is the element of arc length ds. — The arc from $(a, f (a))$ to $(x, f (x))$ . Over an infinitesimal step the curve is virtually straight, so $d s$ , $d x$ , and $d y$ are the sides of a right triangle.

Factoring $d x$ out of the square root,

d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x .

But dy/dx=f'(x), and therefore

ds=\sqrt{1+[f'(x)]^{2}}\,dx.

As $d s$ sweeps along the curve from $(a, f (a))$ to $(b, f (b))$ , we add up all the infinitesimal lengths:

\begin{aligned} \text{length of arc} &= \int_{a}^{b}ds=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx\\ &= \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\,dx. \end{aligned}

Arc Length in Terms of y

If we want to compute the length of the curve $x = h (y)$ between $(h (c), c)$ and $(h (d), d)$ , as in the figure below, we can still start from

d s = \sqrt{(d x)^{2} + (d y)^{2}},

except that now it is $d y$ that we factor out:

\begin{aligned} ds &= \sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ &= \sqrt{[h'(y)]^{2}+1}\ dy \end{aligned}

and

\begin{aligned} \text{length of arc} &= \int_{c}^{d}ds=\int_{c}^{d}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ &= \int_{c}^{d}\sqrt{[h'(y)]^{2}+1}\ dy. \end{aligned}

The graph of a curve x = h(y) with the arc measured from the point (h(c), c) up to the point (h(d), d), and a small right triangle on the curve whose legs are dx and dy and whose hypotenuse is ds. — The curve $x = h (y)$ between $y = c$ and $y = d$ . The same triangle appears, but now it is $d y$ that we factor out of $d s = \sqrt{(d x)^{2} + (d y)^{2}}$ .

Rectifiable Curves

In the above discussion we assumed that the curve is rectifiable; that is, the curve has a finite arc length. It is possible to give an example of a continuous function $y = f (x)$ on $a \leq x \leq b$ whose graph is not rectifiable. One such function is f(x)=\begin{cases} x\sin\dfrac{\pi}{x} & \text{if }x\neq0\\[4pt] 0 & \text{if }x=0, \end{cases}
which is continuous on $[0, 1]$ but is not rectifiable there: it oscillates infinitely often as $x \to 0^{+}$ , and the up-and-down pieces add to an infinite total length even though the graph stays inside a bounded region.

A sufficient condition for the curve $y = f (x)$ to be rectifiable is that f' be bounded, that is, that there be some number $K$ with |f'(x)| for all $x$ between $a$ and $b$ . Then \sqrt{1+[f'(x)]^{2}}<\sqrt{1+K^{2}}, and the arc length integral is at most $(b - a) \sqrt{1 + K^{2}}$ , which is finite.

Worked Examples

In practice the difficulty is never the formula; it is the square root. Almost every arc length problem you can finish by hand is one where 1+(y')^{2} collapses to a perfect square or to something a single substitution can handle. Watch for that in each of the following.

Find the length of the curve $y^{2} = x^{3}$ between the points $(0, 0)$ and $(4, 8)$ .

The curve y squared equals x cubed rising steeply from the origin to the point (4, 8) in the first quadrant. — The arc of $y^{2} = x^{3}$ from $(0, 0)$ to $(4, 8)$ .

Solution

Solving $y^{2} = x^{3}$ for $y$ , we obtain

y = x^{\frac{3}{2}} (for y > 0) \Rightarrow \frac{d y}{d x} = \frac{3}{2} x^{1 / 2} .

The arc length formula yields

\begin{aligned} L &= \int_{0}^{4}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx\\ &= \int_{0}^{4}\sqrt{1+\frac{9}{4}x}\,dx. \end{aligned}

Let $u = 1 + \frac{9}{4} x$ , so that $d u = \frac{9}{4} d x$ and $d x = \frac{4}{9} d u$ . The limits become

x = 0 \Leftrightarrow u = 1, x = 4 \Leftrightarrow u = 10.

Therefore,

\begin{aligned} L &= \int_{1}^{10}\frac{4}{9}\sqrt{u}\ du\\ &= \frac{4}{9}\left(\frac{2}{3}\right)u^{\frac{3}{2}}\Bigg|_{u=1}^{u=10}\\ &= \frac{8}{27}\left(\sqrt{1000}-1\right)\\ &\approx 9.07342. \end{aligned}

Find the length of the curve $y = \ln \sec x$ on $[0, \frac{π}{4}]$ .

The graph of y = ln(sec x), which starts at the origin and increases as x runs from 0 to pi over 4. — Graph of $y = \ln \sec x$ .

Solution

Write $u = \sec x$ . Then by the chain rule

y'=\frac{u'}{u}=(\sec x\tan x)\left(\frac{1}{\sec x}\right)=\tan x,

and the identity $1 + \tan^{2} x = \sec^{2} x$ turns the square root into something we can integrate:

\begin{aligned} L &= \int_{0}^{\frac{\pi}{4}}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx\\ &= \int_{0}^{\frac{\pi}{4}}\sqrt{1+\tan^{2}x}\,dx\\ &= \int_{0}^{\frac{\pi}{4}}\sqrt{\sec^{2}x}\,dx &&{\small(1+\tan^{2}x=\sec^{2}x)}\\ &= \int_{0}^{\frac{\pi}{4}}|\sec x|\,dx\\ &= \int_{0}^{\frac{\pi}{4}}\sec x\,dx &&{\small\left(\sec x>0\text{ for }0\leq x\leq\tfrac{\pi}{4}\right)}\\ &= \Bigg[\ln|\sec x+\tan x|\Bigg]_{x=0}^{x=\frac{\pi}{4}}\\ &= \ln\left|\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\right|-\ln|\sec0+\tan0|\\ &= \ln\left|\frac{1}{\cos\frac{\pi}{4}}+\tan\frac{\pi}{4}\right|-\ln\left|\frac{1}{\cos0}+\tan0\right|\\ &= \ln\left|\frac{2}{\sqrt{2}}+1\right|-\ln|1+0|\\ &= \ln\left(\frac{2}{\sqrt{2}}+1\right)-0\\ &= \ln\left(\sqrt{2}+1\right)\approx0.88137. \end{aligned}

Find the length of the curve $y = a \cosh \frac{x}{a}$ on $[0, a]$ , where $a > 0$ .

Solution

Recall that $\frac{d}{d x} \cosh x = \sinh x$ . Therefore, with $u = \frac{x}{a}$ ,

\begin{aligned} y=a\cosh\underbrace{\frac{x}{a}}_{u}\quad\Rightarrow\quad y' &= au'\sinh u\\ y' &= a\cdot\frac{1}{a}\sinh\frac{x}{a}\\ &= \sinh\frac{x}{a}. \end{aligned}

The identity $\cosh^{2} t - \sinh^{2} t = 1$ makes 1+(y')^{2} a perfect square, and the arc length formula gives

\begin{aligned} L &= \int_{0}^{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx\\ &= \int_{0}^{a}\sqrt{1+\sinh^{2}\frac{x}{a}}\,dx\\ &= \int_{0}^{a}\cosh\frac{x}{a}\,dx &&{\small\left(\cosh^{2}t-\sinh^{2}t=1\text{ and }\cosh t>0\right)}\\ &= a\left.\sinh\frac{x}{a}\right|_{x=0}^{x=a}\\ &= a\left(\sinh1-\sinh0\right)\\ &= a\sinh1\approx1.1752\,a. \end{aligned}

Here we used $\int \cosh t d t = \sinh t + C$ , and hence, substituting $u = \frac{x}{a}$ , so that $x = a u$ and $d x = a d u$ ,

\begin{aligned} \int\cosh\frac{x}{a}\,dx &= \int\cosh u\ \overbrace{a\,du}^{dx}\\ &= a\sinh u+C\\ &= a\sinh\frac{x}{a}+C. \end{aligned}

Differentiating $a \sinh \frac{x}{a}$ returns $a \cdot \frac{1}{a} \cosh \frac{x}{a} = \cosh \frac{x}{a}$ , which confirms the antiderivative.

Find the circumference of the circle $x^{2} + y^{2} = R^{2}$ .

Solution

Using implicit differentiation,

2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} .

Therefore, the element of arc is

\begin{aligned} ds &= \sqrt{(dx)^{2}+(dy)^{2}}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\:dx\\ &= \sqrt{1+\left(-\frac{x}{y}\right)^{2}}\,dx\\ &= \sqrt{\frac{y^{2}+x^{2}}{y^{2}}}\,dx\\ &= \sqrt{\frac{R^{2}}{R^{2}-x^{2}}}\,dx &&{\small\left(y^{2}=R^{2}-x^{2}\right)}\\ &= \frac{R}{\sqrt{R^{2}-x^{2}}}\,dx. \end{aligned}

Because $x$ varies between $- R$ and $R$ , the length of the curve is

s = \int_{- R}^{R} \frac{R}{\sqrt{R^{2} - x^{2}}} d x = \int_{- R}^{R} \frac{1}{\sqrt{1 - \frac{x^{2}}{R^{2}}}} d x .

Let $u = x / R$ . Then $d u = d x / R$ . When $x = R$ we get $u = 1$ , and when $x = - R$ we get $u = - 1$ . Therefore,

\begin{aligned} s &= \int_{-1}^{1}\frac{1}{\sqrt{1-u^{2}}}\overbrace{R\,du}^{dx}\\[4pt] &= R\arcsin u\bigg|_{-1}^{1}\\[4pt] &= R\left(\overbrace{\arcsin1}^{\pi/2}-\overbrace{\arcsin(-1)}^{-\pi/2}\right)\\ &= \pi R. \end{aligned}

Note that this is just the length of a semicircle: as $x$ runs once from $- R$ to $R$ , the formula $y = \sqrt{R^{2} - x^{2}}$ traces only the upper half. The length of the other half is the same by symmetry, so the total circumference of the circle is $2 π R$ .

Discontinuities in dy/dx

Even when $d y / d x$ is not defined at some points on the curve $y = f (x)$ , it is possible that $d x / d y$ is defined everywhere on this curve. An example of such a situation is when the curve $y = f (x)$ has a vertical tangent. In such cases, to compute the arc length of the curve $y = f (x)$ , we can express $x$ in terms of $y$ and write

d s = \sqrt{(d x)^{2} + (d y)^{2}}

\Rightarrow \frac{d s}{d y} = \sqrt{{(\frac{d x}{d y})}^{2} + 1} .

Find the length of the curve $y = x^{2 / 3}$ from $x = 0$ to $x = 8$ .

The graph of y = x to the two-thirds power, with a sharp cusp and a vertical tangent at the origin, rising to the right to the point (8, 4). — The curve $y = x^{2 / 3}$ from $x = 0$ to $x = 8$ . The tangent at the origin is vertical, so $d y / d x$ is undefined there.

Solution

Differentiating directly,

y=x^{2/3}\quad\Rightarrow\quad y'=\frac{2}{3}x^{-1/3}.

Because y' is not defined at $x = 0$ , we cannot find the length of the curve this way: the tangent line there is vertical. However, if we express $x$ in terms of $y$ , we find

x = y^{3 / 2} \Rightarrow \frac{d x}{d y} = \frac{3}{2} y^{1 / 2},

which is perfectly well behaved on the whole interval. The endpoints transfer as

x = 0 \Leftrightarrow y = 0, x = 8 \Leftrightarrow y = 4.

The length of the curve is

\begin{aligned} s &= \int_{0}^{4}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ &= \int_{0}^{4}\sqrt{\frac{9}{4}y+1}\ dy\\ &= \frac{1}{2}\int_{0}^{4}\sqrt{9y+4}\ dy\\ &= \frac{1}{2}\int_{4}^{40}\sqrt{u}\ \frac{du}{9} &&{\small(u=9y+4\Rightarrow du=9\,dy)}\\ &= \frac{1}{18}\left[\frac{2}{3}u^{3/2}\right]_{4}^{40}\\ &= \frac{1}{27}\left(40^{3/2}-8\right)\\ &= \frac{1}{27}\left(8\sqrt{10^{3}}-8\right)\\ &= \frac{8}{27}\left(\sqrt{1000}-1\right)\\ &\approx 9.07342. \end{aligned}

This is the same number we obtained in the first example, and that is no accident: it is the same arc. The curve $y = x^{2 / 3}$ is exactly the curve $y = x^{3 / 2}$ (that is, $y^{2} = x^{3}$ ) reflected in the line $y = x$ , and the reflection carries the arc from $(0, 0)$ to $(4, 8)$ onto the arc from $(0, 0)$ to $(8, 4)$ . Reflecting a curve does not change its length, and indeed the two computations reduce to the identical integral $\int_{0}^{4} \sqrt{1 + \frac{9}{4} t} d t = \frac{8}{27} (\sqrt{1000} - 1) \approx 9.07342$ . The only difference is that in the first example the variable of integration was $x$ and here it is $y$ , which is precisely what rescued us from the vertical tangent.

The Arc Length Function

The length of the curve $y = f (x)$ from $(a, f (a))$ to $(b, f (b))$ is

\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\,dx.

Let $s (x)$ be a function that measures the arc length from $(a, f (a))$ to a variable point $(x, f (x))$ . To find the formula for $s (x)$ , we just need to replace $b$ in the above formula with $x$ :

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{s(x)=\int_{a}^{x}\sqrt{1+[f'(t)]^{2}}\ dt}

Because we can denote the variable of integration by any arbitrary letter, and because $x$ appears as the upper limit of integration, we have replaced the variable of integration by $t$ so that $x$ does not have two different meanings. This function is called the arc length function.

By the Fundamental Theorem of Calculus, \frac{ds}{dx}=\sqrt{1+[f'(x)]^{2}}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}
or, multiplying through by $d x$ , back to where we started:
$d s = \sqrt{(d x)^{2} + (d y)^{2}} .$
Note that $s (x)$ is an increasing function. To prove this property, we need to show s'(x)>0: s'(x)=\sqrt{1+[f'(x)]^{2}}>0,
which holds for every $x$ , because the quantity under the radical is at least $1$ . This matches intuition: as you move along the curve, the distance you have travelled never decreases.

Find the arc length function for the curve $y = \frac{1}{2} x^{2} - \frac{1}{4} \ln x$ taking $(1, 0.5)$ as the starting point. Then, using this function, find the arc length along this curve from $(1, 0.5)$ to $(e, 0.5 e^{2} - 0.25)$ .

Solution

First differentiate:

f(x)=\frac{1}{2}x^{2}-\frac{1}{4}\ln x\quad\Rightarrow\quad f'(x)=x-\frac{1}{4x}.

The middle term of the square is $- 2 \cdot x \cdot \frac{1}{4 x} = - \frac{1}{2}$ , so adding $1$ turns the expression into a perfect square:

\begin{aligned} \sqrt{1+[f'(x)]^{2}} &= \sqrt{1+\left(x-\frac{1}{4x}\right)^{2}}\\ &= \sqrt{1+x^{2}-\frac{1}{2}+\frac{1}{16x^{2}}}\\ &= \sqrt{\left(x+\frac{1}{4x}\right)^{2}}\\ &= x+\frac{1}{4x} &&{\small(\text{because }x>0)}. \end{aligned}

Therefore, the arc length function is

\begin{aligned} s(x) &= \int_{1}^{x}\sqrt{1+[f'(t)]^{2}}\ dt\\ &= \int_{1}^{x}\left(t+\frac{1}{4t}\right)dt\\ &= \left[\frac{1}{2}t^{2}+\frac{1}{4}\ln t\right]_{1}^{x}\\ &= \frac{1}{2}x^{2}+\frac{1}{4}\ln x-\frac{1}{2}-\frac{1}{4}\underbrace{\ln1}_{=0}\\ &= \frac{1}{2}x^{2}+\frac{1}{4}\ln x-\frac{1}{2}. \end{aligned}

As a check, $s (1) = \frac{1}{2} + 0 - \frac{1}{2} = 0$ , as it must be at the starting point. The arc length along this curve from $(1, \frac{1}{2})$ to $(e, 0.5 e^{2} - 0.25)$ is

\begin{aligned} s(e) &= \frac{1}{2}e^{2}+\frac{1}{4}\underbrace{\ln e}_{=1}-\frac{1}{2}\\ &= \frac{e^{2}}{2}-\frac{1}{4}\\ &\approx 3.44453. \end{aligned}

Exercises

In each problem, look first for the algebra that makes 1+(y')^{2} a perfect square. When the derivative blows up, integrate with respect to $y$ instead, as in the $y = x^{2 / 3}$ example above.

Find the length of the curve $y = \frac{2}{3} x^{3 / 2}$ from $x = 0$ to $x = 8$ .

Solution

Here y'=x^{1/2}, so 1+(y')^{2}=1+x, and the square root disappears at once:

\begin{aligned} L &= \int_{0}^{8}\sqrt{1+x}\ dx\\ &= \frac{2}{3}(1+x)^{3/2}\bigg|_{0}^{8}\\ &= \frac{2}{3}\left(27-1\right)\\ &= \frac{52}{3}\approx17.3333. \end{aligned}

Find the length of the curve $y = \frac{x^{3}}{6} + \frac{1}{2 x}$ from $x = 1$ to $x = 2$ .

Solution

Differentiating, y'=\dfrac{x^{2}}{2}-\dfrac{1}{2x^{2}}. The cross term in (y')^{2} is $- 2 \cdot \frac{x^{2}}{2} \cdot \frac{1}{2 x^{2}} = - \frac{1}{2}$ , so

1+(y')^{2}=1+\frac{x^{4}}{4}-\frac{1}{2}+\frac{1}{4x^{4}}=\frac{x^{4}}{4}+\frac{1}{2}+\frac{1}{4x^{4}}=\left(\frac{x^{2}}{2}+\frac{1}{2x^{2}}\right)^{2}.

Since $x > 0$ , the square root is $\frac{x^{2}}{2} + \frac{1}{2 x^{2}}$ , and

\begin{aligned} L &= \int_{1}^{2}\left(\frac{x^{2}}{2}+\frac{1}{2x^{2}}\right)dx\\ &= \left[\frac{x^{3}}{6}-\frac{1}{2x}\right]_{1}^{2}\\ &= \left(\frac{8}{6}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{2}\right)\\ &= \frac{4}{3}-\frac{1}{4}+\frac{1}{3}\\ &= \frac{17}{12}\approx1.41667. \end{aligned}

Notice that the antiderivative is the original function with the sign of its second term flipped. That is a recurring feature of this family of curves.

Find the length of the curve $y = \frac{x^{2}}{4} - \frac{\ln x}{2}$ from $x = 1$ to $x = e$ .

Solution

We have y'=\dfrac{x}{2}-\dfrac{1}{2x}, whose square has cross term $- \frac{1}{2}$ . Hence

1+(y')^{2}=1+\frac{x^{2}}{4}-\frac{1}{2}+\frac{1}{4x^{2}}=\left(\frac{x}{2}+\frac{1}{2x}\right)^{2},

and since $x > 0$ ,

\begin{aligned} L &= \int_{1}^{e}\left(\frac{x}{2}+\frac{1}{2x}\right)dx\\ &= \left[\frac{x^{2}}{4}+\frac{\ln x}{2}\right]_{1}^{e}\\ &= \left(\frac{e^{2}}{4}+\frac{1}{2}\right)-\left(\frac{1}{4}+0\right)\\ &= \frac{e^{2}}{4}+\frac{1}{4}\approx2.09726. \end{aligned}

Find the length of the curve $y = \ln \cos x$ from $x = 0$ to $x = \frac{π}{3}$ .

Solution

Here y'=\dfrac{-\sin x}{\cos x}=-\tan x, so 1+(y')^{2}=1+\tan^{2}x=\sec^{2}x. On $[0, \frac{π}{3}]$ we have $\sec x > 0$ , so $\sqrt{\sec^{2} x} = \sec x$ and

\begin{aligned} L &= \int_{0}^{\pi/3}\sec x\,dx\\ &= \Big[\ln|\sec x+\tan x|\Big]_{0}^{\pi/3}\\ &= \ln\left(2+\sqrt{3}\right)-\ln(1+0)\\ &= \ln\left(2+\sqrt{3}\right)\approx1.31696. \end{aligned}

This is the mirror image of the example $y = \ln \sec x$ solved in the text: replacing $\sec$ by $\cos$ only changes the sign of $y$ , and reflecting a curve in the $x$ -axis does not change its length.

Find the length of the catenary $y = \cosh x$ from $x = - 1$ to $x = 1$ .

Solution

Since y'=\sinh x and $\cosh^{2} x - \sinh^{2} x = 1$ , we get 1+(y')^{2}=\cosh^{2}x. Because $\cosh x > 0$ everywhere,

\begin{aligned} L &= \int_{-1}^{1}\cosh x\,dx\\ &= \Big[\sinh x\Big]_{-1}^{1}\\ &= \sinh1-\sinh(-1)=2\sinh1\\ &= e-\frac{1}{e}\approx2.35040. \end{aligned}

The chord joining $(- 1, \cosh 1)$ to $(1, \cosh 1)$ has length $2$ , so the curve is only about $17.5 %$ longer than the straight line between its endpoints.

Find the length of the curve $x = \frac{y^{3}}{6} + \frac{1}{2 y}$ from $y = 1$ to $y = 3$ .

Solution

The curve is given as a function of $y$ , so we use the second form of the formula. Here

\frac{d x}{d y} = \frac{y^{2}}{2} - \frac{1}{2 y^{2}},

and exactly as in the second exercise,

{(\frac{d x}{d y})}^{2} + 1 = {(\frac{y^{2}}{2} + \frac{1}{2 y^{2}})}^{2} .

Since $y > 0$ on the interval,

\begin{aligned} L &= \int_{1}^{3}\left(\frac{y^{2}}{2}+\frac{1}{2y^{2}}\right)dy\\ &= \left[\frac{y^{3}}{6}-\frac{1}{2y}\right]_{1}^{3}\\ &= \left(\frac{27}{6}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{2}\right)\\ &= \frac{13}{3}+\frac{1}{3}=\frac{14}{3}\approx4.66667. \end{aligned}

Find the length of the curve $y = x^{2 / 3}$ from $x = - 1$ to $x = 1$ .

Solution

The derivative y'=\frac{2}{3}x^{-1/3} is undefined at $x = 0$ , where the graph has a cusp with a vertical tangent, so we integrate with respect to $y$ . The curve is symmetric about the $y$ -axis, so the total length is twice the length of the right-hand half.

On the right half, $y = x^{2 / 3}$ gives $x = y^{3 / 2}$ with $\frac{d x}{d y} = \frac{3}{2} y^{1 / 2}$ , and $x$ runs from $0$ to $1$ as $y$ runs from $0$ to $1$ . Hence

\begin{aligned} \frac{L}{2} &= \int_{0}^{1}\sqrt{\frac{9}{4}y+1}\ dy\\ &= \frac{4}{9}\cdot\frac{2}{3}\left(1+\frac{9}{4}y\right)^{3/2}\bigg|_{0}^{1}\\ &= \frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right]\\ &= \frac{8}{27}\left(\frac{13\sqrt{13}}{8}-1\right)=\frac{13\sqrt{13}-8}{27}. \end{aligned}

Therefore,

L = \frac{26 \sqrt{13} - 16}{27} \approx 2.87942 .

Set up the integral for the length of one arch of $y = \sin x$ , from $x = 0$ to $x = π$ . Explain why the techniques of this section do not evaluate it, and give the value to four decimal places.

Solution

Since y'=\cos x, the arc length integral is

L = \int_{0}^{π} \sqrt{1 + \cos^{2} x} d x .

The integrand is where the method stops. Unlike $1 + \tan^{2} x = \sec^{2} x$ or $1 + \sinh^{2} t = \cosh^{2} t$ , the expression $1 + \cos^{2} x$ is not a perfect square and no substitution reduces it to one: this is an elliptic integral, and it has no antiderivative expressible in terms of elementary functions. Setting up the integral correctly is the whole of the calculus here; the number must come from a numerical method.

Applying Simpson's rule on $[0, \frac{π}{2}]$ and doubling (the arch is symmetric about $x = \frac{π}{2}$ ) gives

L \approx 3.8202 .

As a sanity check, the arch rises $1$ unit and spans $π \approx 3.1416$ units, so a length slightly under $4$ is exactly what we should expect.

Find the length of the parabola $y = x^{2}$ from $(0, 0)$ to $(2, 4)$ . (You will need $\int \sqrt{1 + u^{2}} d u = \frac{1}{2} u \sqrt{1 + u^{2}} + \frac{1}{2} \ln (u + \sqrt{1 + u^{2}}) + C$ .)

Solution

Here y'=2x, so

L = \int_{0}^{2} \sqrt{1 + 4 x^{2}} d x .

Put $u = 2 x$ , so that $d u = 2 d x$ and the limits become $u = 0$ and $u = 4$ :

L = \frac{1}{2} \int_{0}^{4} \sqrt{1 + u^{2}} d u .

Using the stated antiderivative,

\begin{aligned} L &= \frac{1}{2}\left[\frac{1}{2}u\sqrt{1+u^{2}}+\frac{1}{2}\ln\left(u+\sqrt{1+u^{2}}\right)\right]_{0}^{4}\\ &= \frac{1}{4}\left[4\sqrt{17}+\ln\left(4+\sqrt{17}\right)\right]-0\\ &= \sqrt{17}+\frac{1}{4}\ln\left(4+\sqrt{17}\right)\\ &\approx 4.64678. \end{aligned}

The chord from $(0, 0)$ to $(2, 4)$ has length $\sqrt{20} \approx 4.47214$ , and the arc is a little longer, as it must be.

Find the arc length function $s (x)$ for the curve $y = \frac{x^{3}}{3} + \frac{1}{4 x}$ with starting point $(1, \frac{7}{12})$ , and use it to find the length of the arc from $x = 1$ to $x = 3$ .

Solution

Differentiating, f'(x)=x^{2}-\dfrac{1}{4x^{2}}, whose square has cross term $- 2 \cdot x^{2} \cdot \frac{1}{4 x^{2}} = - \frac{1}{2}$ . Therefore

1+[f'(x)]^{2}=x^{4}+\frac{1}{2}+\frac{1}{16x^{4}}=\left(x^{2}+\frac{1}{4x^{2}}\right)^{2},

and for $x > 0$ the square root is $x^{2} + \frac{1}{4 x^{2}}$ . The arc length function is

\begin{aligned} s(x) &= \int_{1}^{x}\left(t^{2}+\frac{1}{4t^{2}}\right)dt\\ &= \left[\frac{t^{3}}{3}-\frac{1}{4t}\right]_{1}^{x}\\ &= \frac{x^{3}}{3}-\frac{1}{4x}-\left(\frac{1}{3}-\frac{1}{4}\right)\\ &= \frac{x^{3}}{3}-\frac{1}{4x}-\frac{1}{12}. \end{aligned}

Check: $s (1) = \frac{1}{3} - \frac{1}{4} - \frac{1}{12} = 0$ , and s'(x)=x^{2}+\frac{1}{4x^{2}}>0, so $s$ increases, as an arc length function must. Finally

s (3) = 9 - \frac{1}{12} - \frac{1}{12} = 9 - \frac{1}{6} = \frac{53}{6} \approx 8.83333 .

Find the total length of the astroid $x^{2 / 3} + y^{2 / 3} = a^{2 / 3}$ , where $a > 0$ .

Solution

The curve is symmetric in both axes, so the total length is four times the length of the quarter lying in the first quadrant, where

y = {(a^{2 / 3} - x^{2 / 3})}^{3 / 2}, 0 \leq x \leq a .

By the chain rule,

y'=\frac{3}{2}\left(a^{2/3}-x^{2/3}\right)^{1/2}\cdot\left(-\frac{2}{3}x^{-1/3}\right)=-\frac{\left(a^{2/3}-x^{2/3}\right)^{1/2}}{x^{1/3}},

so that

1+(y')^{2}=1+\frac{a^{2/3}-x^{2/3}}{x^{2/3}}=\frac{x^{2/3}+a^{2/3}-x^{2/3}}{x^{2/3}}=\frac{a^{2/3}}{x^{2/3}}.

Taking the square root (both $a$ and $x$ are positive),

\sqrt{1+(y')^{2}}=\frac{a^{1/3}}{x^{1/3}}.

This blows up as $x \to 0^{+}$ , but the resulting improper integral converges because the exponent $\frac{1}{3}$ is less than $1$ :

\begin{aligned} \frac{L}{4} &= \int_{0}^{a}a^{1/3}x^{-1/3}\,dx\\ &= a^{1/3}\cdot\frac{3}{2}x^{2/3}\bigg|_{0}^{a}\\ &= \frac{3}{2}a^{1/3}a^{2/3}=\frac{3}{2}a. \end{aligned}

Therefore the total length of the astroid is

L = 4 \cdot \frac{3}{2} a = 6 a .

For comparison, the circle of the same radius $a$ has circumference $2 π a \approx 6.283 a$ , so the astroid, pinched into four cusps, is slightly shorter.

Find the length of the curve $y = \ln x$ from $x = 1$ to $x = \sqrt{3}$ .

Solution

Since y'=\dfrac{1}{x},

L = \int_{1}^{\sqrt{3}} \sqrt{1 + \frac{1}{x^{2}}} d x = \int_{1}^{\sqrt{3}} \frac{\sqrt{x^{2} + 1}}{x} d x,

using $x > 0$ to write $\sqrt{x^{2}} = x$ . There is no perfect square here, so substitute $u = \sqrt{x^{2} + 1}$ . Then $u^{2} = x^{2} + 1$ and $u d u = x d x$ , so $d x = \frac{u d u}{x}$ and

\frac{\sqrt{x^{2} + 1}}{x} d x = \frac{u}{x} \cdot \frac{u d u}{x} = \frac{u^{2}}{x^{2}} d u = \frac{u^{2}}{u^{2} - 1} d u = (1 + \frac{1}{u^{2} - 1}) d u .

The limits are $x = 1 \Rightarrow u = \sqrt{2}$ and $x = \sqrt{3} \Rightarrow u = 2$ . Since $\frac{1}{u^{2} - 1} = \frac{1}{2} (\frac{1}{u - 1} - \frac{1}{u + 1})$ ,

\begin{aligned} L &= \left[u+\frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|\right]_{\sqrt{2}}^{2}\\ &= \left(2+\frac{1}{2}\ln\frac{1}{3}\right)-\left(\sqrt{2}+\frac{1}{2}\ln\frac{\sqrt{2}-1}{\sqrt{2}+1}\right). \end{aligned}

Now $\frac{\sqrt{2} - 1}{\sqrt{2} + 1} = {(\sqrt{2} - 1)}^{2}$ , so $\frac{1}{2} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \ln (\sqrt{2} - 1) = - \ln (\sqrt{2} + 1)$ . Therefore

\begin{aligned} L &= 2-\sqrt{2}-\frac{1}{2}\ln3+\ln\left(\sqrt{2}+1\right)\\ &= 2-\sqrt{2}+\ln\frac{\sqrt{2}+1}{\sqrt{3}}\\ &\approx 0.91785. \end{aligned}

The straight-line distance between $(1, 0)$ and $(\sqrt{3}, \ln \sqrt{3})$ is $\sqrt{{(\sqrt{3} - 1)}^{2} + {(\ln \sqrt{3})}^{2}} \approx 0.91523$ , so the arc is only marginally longer, which fits a curve this gently bent.

The Differential of Arc Length

Arc Length in Terms of y

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Discontinuities in dy/dx

The Arc Length Function

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